Modern Physics Third Edition RAYMOND A. SERWAY CLEMENT J. MOSES CURT A. MOYER
1 RELATIVITY 1.1 Special Relativity 1.2 The Principle of Relativity, The Speed of Light 1.3 The Michelson Morley Experiment, Details of the Michelson Morley Experiment. 1.4 Postulates of Special Relativity. 1.5 Consequences of Special Relativity, Simultaneity and the Relativity of Time, Time Dilation Length Contraction. 1.6 The Lorentz Transformation,Lorentz Velocity Transformation
1.1 SPECIAL RELATIVITY Classical mechanics Newton s laws: 1. Inertia: Any object moves with constant velocity as long as no net force acts upon it 2. Action: any object experiences acceleration in presence of a net force: F = ma 3. Reaction If force F acts upon an object, F acts upon the object : where the force originates.
* To describe processes in nature one needs Reference Frame. Reference Frame = Coordinate system to measure position + Clock to measure time. Inertial system: Reference frame where Newton s 1. law is applied. *Newtonian mechanics valid for object with velocity v<<c. (c: the speed of light)
Light waves and other forms of electromagnetic radiation travel through free space at the speed c = 3.00 x10 8 m/s. As we shall see in this chapter, the speed of light sets an upper limit for the speeds of particles, waves, and the transmission of information.
Newtonian mechanics (Classical Mechanics) deals with objects that move at speeds much less than that of light, v<<c. Or objects in our every day life.
Although Newtonian mechanics works very well at low speeds, it fails when applied to particles whose speeds approach that of light Experimentally, Physicists try to accelerate an electron through a large electric potential difference. For example, it is possible to accelerate an electron to a speed of 0.99c by using a potential difference of several million volts. According to Newtonian mechanics, if the potential difference (as well as the corresponding energy) is increased by a factor of 4, then the speed of the electron should be doubled to 1.98c
However, experiments show that the speed of the electron as well as the speeds of all other particles in the universe always remains less than the speed of light!!!
*In 1905, at the age of 26, Albert Einstein published his special theory of relativity. Regarding the theory, Einstein wrote, (The relativity theory arose from necessity, from serious and deep contradictions in the old theory from which there seemed no escape. The strength of the new theory lies in the consistency and simplicity with which it solves all these difficulties, using only a few very convincing assumptions)
The General Theory What is the general theory of relativity (Einstein 1915)? It describes the relationships between gravity and the geometrical structure of space and time. Remarkable results include: light rays are affected by gravity, and the big bang theory (the universe is continually expanding) The general theory of relativity concerns with accelerating frames of reference Special theory of relativity, on the contrary, is only concerned with inertial frames of reference, that is, frames moving with constant velocities (no acceleration)
postulates of special theory of relativity The laws of physics are the same in all reference systems that move uniformly with respect to one another. That is, basic laws such as_σf = dp/dt have the same mathematical form for all observers moving at constant velocity with respect to one another.
The speed of light in vacuum is always measured to be 3 x10 8 m/s, and the measured value is independent of the motion of the observer or of the motion of the source of light. That is, the speed of light is the same for all observers moving at constant velocities. The Speed of light is a universal constant.
1.2 THE PRINCIPLE OF RELATIVITY Recall from your studies in mechanics that Newton s laws are valid in inertial frames of reference. inertial frame: is one in which an object subjected to no forces moves in a straight line at constant speed thus the name inertial frame because an object observed from such a frame obeys Newton s first law, the law of inertia. Furthermore, any frame or system moving with constant velocity with respect to an inertial system must also be an inertial system. Thus there is no single, preferred inertial frame for applying Newton s laws
According to the principle of Newtonian relativity, the laws of mechanics must be the same in all inertial frames of reference For example
Galilean transformations The observer in the truck throws a ball straight up There is a stationary observer on the ground It appears to move in a vertical path Views the path of the ball thrown to be a parabola The law of gravity and equations of motion under uniform acceleration are obeyed The ball has a velocity to the right equal to the velocity of the truck
The observer in the truck throws a ball straight up It appears to move in a vertical path The law of gravity and equations of motion under uniform acceleration are obeyed There is a stationary observer on the ground Views the path of the ball thrown to be a parabola, The ball has a velocity to the right equal to the velocity of the truck
Indeed, one of the firm philosophical principles of modern science is that all observers are equivalent and that the laws of nature must take the same mathematical form for all observers. Laws of physics that exhibit the same mathematical form for observers with different motions at different locations are said to be covariant.
Definition of Event: An event is determined by the location where and by the time at which occurs, i.e by (t, r) Two events are simultaneous if the time of the first event equals the time of the second event In Classical mechanics time is absolute, i.e two simultaneous events will be simultaneous in any IRF.
In order to show this let us make the following illustration:
Consider two inertial systems or frames S and S', as in Figure 1.2. The frame S' moves with a constant velocity v along the xx' axes, where v is measured relative to the frame S. Clocks in S and S' are synchronized, and the origins of S and S' coincide at t = t'= 0.
We assume that a point event, a physical phenomenon such as a lightbulb flash, occurs at the point P. An observer in the system S would describe the event with space time coordinates (x, y, z, t), whereas an observer in S would use (x', y', z', t') to describe the same event. As we can see from Figure 1.2, these coordinates are related by the equations
These equations constitute what is known as a Galilean transformation of coordinates
Note that the fourth coordinate, time, is assumed to be the same in both inertial frames. That is, in classical mechanics, all clocks run at the same rate regardless of their velocity, so that the time at which an event occurs for an observer in S is the same as the time for the same event in S'. Consequently, the time interval between two successive events should be the same
for both observers. Although this assumption may seem obvious, it turns out to be incorrect when treating situations in which v is comparable to the speed of light. In fact, this point represents one of the most profound differences between Newtonian concepts and the ideas contained in Einstein s theory of relativity
Now suppose two events are separated by a distance dx and a time interval dt as measured by an observer in S. It follows from Equation 1.1 that the corresponding displacement dx measured by an observer in S' is given by dx'= dx - v dt, where dx is the displacement measured by an observer in S. Because dt = dt ', we find that
or where u x and u' x are the instantaneous velocities of the object relative to S and S', respectively. This result, which is called the Galilean addition law for velocities (or Galilean velocity transformation), is used in everyday observations and is consistent with our intuitive notions of time and space
To obtain the relation between the accelerations measured by observers in S and S', we take a derivative of Equation 1.2 with respect to time and use the results that dt = dt' and v is constant
Galilean Space-Time Transformation t t z z y y vt x x ' ' ' ' Coordinates: Velocities: z z y y x x u u u u v u v dt dx dt dx u ' ' ' - ' ' Accelerations: z z y y x x a a a a a dt x d dt x d a ' ' 2 2 2 2 ' ' ' Newton s Laws involving accelerations are invariant with respect to Galilean transformations!
EXAMPLE 1.1 Assume that Newton s law F x = ma x has been shown to hold by an observer in an inertial frame S. Show that Newton s law also holds for an observer in S' or is covariant under the Galilean transformation, that is, has the form F' x = m'a' x. Note that inertial mass is an invariant quantity in Newtonian dynamics.
Solution : Starting with the established law F x = ma x, we use the Galilean transformation a' x =a x and the fact that m' = m to obtain F x = m'a' x. If we now assume that Fx depends only on the relative positions of m and the particles interacting with m, that is, F x = f(x2 - x1, x3 - x1,...), then F x = F ' x, because the x s are invariant quantities. Thus we find F 'x = m'a' x and establish the covariance of Newton s second law in this simple case
In Relativistic Mechanics time is relative * We will show that in Relativistic Mechanics time is relative, i.e clocks record the passing of time differently in different inertial reference frames Example: consider a light signal emitted from A 0 Event 1 =signal arrives A 1 Event 2 =signal arrives A 2 In IRF S events 1 and 2 are simultaneous if A 0 is at rest an A 0 A 1 =A 0 A 2, i.e t 1 =t 2 In the IRF S which moves with v towards A 0 evidently t 0 t 0 An d the events are not simultaneous anymore.
'S S v 'O o... 1A 0 A 2 A c c
The Speed of Light Maxwell in the 1860s showed that the speed of light in free space was given by *Physicists of the late 1800s were certain that light waves (like familiar sound and water waves) required a definite medium in which to move, called the ether *and that the speed of light was c only with respect to the ether or a frame fixed in the ether called the ether frame. In any other frame moving at speed v relative to the ether frame
*the Galilean addition law was expected to hold. Thus, the speed of light in this other frame was expected to be c - v for light traveling in the same direction as the frame, c + v for light traveling opposite to the frame, and in between these two values for light moving in an arbitrary direction with respect to the moving frame If v is the velocity of the ether relative to the Earth, then the speed of light should have its maximum value, c + v, when propagating downwind, as shown in Figure 1.3a
Likewise, the speed of light should have its minimum value, c - v, when propagating upwind, as in Figure 1.3b, and an intermediate value, (c 2 -v 2 ) 1/2, in the direction perpendicular to the ether wind, as in Figure 1.3c If the Sun is assumed to be at rest in the ether, then the velocity of the ether wind would be equal to the orbital velocity of the Earth around the Sun, which has a magnitude of about 3 x 10 4 m/s compared to c = 3 x 10 8 m/s. Thus, the change in the speed of light would be about 1 part in 10 4 for measurements in the upwind or downwind directions, and changes of this size should be detectable
1.3 THE MICHELSON MORLEY EXPERIMENT The aim of this experiment is to prove that Ether hypothesis is false The Michelson interferometer was designed to detect small changes in the speed of light with motion of an observer through the ether. One of the arms of the interferometer was aligned along the direction of the motion of the Earth through the ether. The Earth moving through the ether would be equivalent to the ether flowing past the Earth in the opposite direction with speed v, **This ether wind blowing in the opposite direction should cause the speed of light measured in the Earth s frame of reference to be c - v as it approaches the mirror M 2 in Figure 1.4 and c + v after reflection. The speed v is the speed of the Earth through space, and hence the speed of the ether wind, and c is the speed of light in the ether frame. The two beams of light reflected from M 1 and M 2 would recombine, and an interference pattern consisting of alternating dark and bright bands, or fringes, would be formed
* During the experiment, the interference pattern was observed while the interferometer was rotated through an angle of 90. This rotation would change the speed of the ether wind along the direction of the arms of the interferometer. The effect of this rotation should hav been to cause the fringe pattern to shift slightly but measurably. * Measurements failed to show any change in the interference pattern * when the ether wind was expected to have changed direction and magnitude, but the results were always the same: No fringe shift of the magnitude required was ever observed * The negative results of the Michelson Morley experiment not only meant that the speed of light does not depend on the direction of light propagation but also contradicted the ether hypothesis. The negative results also meant that it was impossible to measure the absolute velocity of the Earth with respect to the ether frame
Details of the Michelson Morley Experiment let us assume that the interferometer shown in Figure 1.4 has two arms of equal length L. First consider the beam traveling parallel to the direction of the ether wind, which is taken to be horizontal in Figure 1.4. According to Newtonian mechanics, as the beam moves to the right, its speed is reduced by the wind and its speed with respect to the Earth is c - v. On its return journey, as the light beam moves to the left downwind, its speed with respect to the Earth is c + v. Thus, the time of travel to the right is L/(c - v), and the time of travel to the left is L/(c + v). The total time of travel for the round-trip along the horizontal path is Now consider the light beam traveling perpendicular to the wind, as shown in Figure 1.4. Because the speed of the beam relative to the Earth is (c 2 - v 2 )1/2 in this case (see Fig. 1.3c), the time of travel for each half of this trip is L/(c 2 - v 2 ) 1/2, and the total time of travel for the round-trip is
Thus, the time difference between the light beam traveling horizontally and the beam traveling vertically is Because v 2 /c 2 << 1, this expression can be simplified by using the following binomial expansion after dropping all terms higher than second order In our case, x = v 2 /c 2, and we find *The two light beams start out in phase and return to form an interference pattern
*Let us assume that the interferometer is adjusted for parallel fringes and that a telescope is focused on one of these fringes. The time difference between the two light beams gives rise to a phase difference between the beams, producing the interference fringe pattern when they combine at the position of the telescope. A difference in the pattern (Fig. 1.6) should be detected by rotating the interferometer through 90 0 in a horizontal plane, such that the two beams exchange roles. This results in a net time difference of twice that given by Equation 1.4. The path difference corresponding to this time difference is The corresponding fringe shift is equal to this path difference divided by the wavelength of light,, because a change in path of 1 wavelength corresponds to a shift of 1 fringe
In the experiments by Michelson and Morley, each light beam was reflected by mirrors many times to give an increased effective path length L of about 11 m. Using this value, and taking v to be equal to 3 _ 104 m/s, the speed of the Earth about the Sun, gives a path difference of
This extra distance of travel should produce a noticeable shift in the fringe pattern. Specifically, using light of wavelength 500 nm, we find a fringe shift for rotation through 90 0 of The precision instrument designed by Michelson and Morley had the capability of detecting a shift in the fringe pattern as small as 0.01 fringe. Howeve they detected no shift in the fringe pattern. Since then, the experiment has been repeated many times by various scientists under various conditions, and no fringe shift has ever been detected. Thus, it was concluded that one cannot detect the motion of the Earth with respect to the ether
1.4 POSTULATES OF SPECIAL RELATIVITY 1- we noted the impossibility of measuring the speed of the ether with respect to the Earth 2- the failure of the Galilean velocity transformation in the case of light *Einstein based his special theory of relativity on two postulates: 1. The Principle of Relativity: All the laws of physics have the same form in all inertial reference frames. 2. The Constancy of the Speed of Light: The speed of light in vacuum has the same value, c = 3.00 x10 8 m/s, in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light
Time Dilation: The fact that observers in different inertial frames always measure different time intervals between a pair of events can be illustrated in another way.
*A mirror is fixed to the ceiling of the vehicle, and observer O', at rest in this system, holds a laser a distance d below the mirror (event 1): the laser emits a pulse of light directed toward the mirror (event 2): after reflecting from the mirror, the pulse arrives back at the laser Aim: to measure the time interval t ' between these two events *Observer O' carries a clock, C *the time it takes to travel from O' to the mirror and back can be found from the definition of speed This time interval t ' measured by O', who, remember, is at rest in the moving vehicle requires only a single clock, C', in this reference frame
*Now consider the same set of events as viewed by observer O in a second frame (Fig. 1.10b). According to this observer, the mirror and laser are moving to the right with a speed v, and as a result, the sequence of events appears different to this observer *By the time the light from the laser reaches the mirror, the mirror has moved to the right a distance v t/2, where t is the time interval required for the light pulse to travel from O' to the mirror and back as measured by O. *In other words, O concludes that, because of the motion of the vehicle, if the light is to hit the mirror, it must leave the laser at an angle with respect to the vertical direction. Comparing Figures 1.10a and 1.10b, we see that the light must travel farther in (b) than in (a). (Note that neither observer knows that he is moving. Each is at rest in his own inertial frame.)
*According to the second postulate of special relativity, both observers must measure c for the speed of light. Because the light travels farther according to O. *it follows that the time interval t measured by O is longer than the time interval t ' measured by O'. *To obtain a relationship between t and t ', it is convenient to use the right triangle shown in Figure 1.10c
Solving for t
Because t '= 2d/c, we can express Equation 1.7 as where = (1 - v 2 /c 2 ) -1/2. Because is always greater than unity, this result says that the time interval t measured by the observer moving with respect to the clock is longer than the time interval t' measured by the observer at rest with respect to the clock. This effect is known as time dilation time dilation: A moving clock runs slower than a clock at rest by a factor of *The time interval t ' in Equation 1.8 is called the proper time. In general, *proper time, denoted t p, is defined as the time interval between two events as measured by an observer who sees the events occur at the same point in space
*In our case, observer O' measures the proper time. That is, proper time is always the time measured by an observer moving along with the clock. As an aid in solving problems it is convenient to express Equation 1.8 in terms of the proper time interval, t p, as EXAMPLE 1.2 The period of a pendulum is measured to be 3.0 s in the rest frame of the pendulum. What is the period of the pendulum when measured by an observer moving at a speed of 0.95c with respect to the pendulum? Solution In this case, the proper time is equal to 3.0 s. From the point of view of the observer, the pendulum is moving at 0.95c past her. Hence the pendulum is an example of a moving clock. Because a moving clock runs slower than a stationary clock by, Equation 1.8 gives
That is, a moving pendulum slows down or takes longer to complete one period
Length Contraction A spaceship travels between two points A and B. Two observers measure the distance between A and B The first observer is on the earth (reference S) The second observer is in the spaceship (reference S') which is moving from A to B with constant speed v with respect to reference frame S (earth). What is the time interval to travel from A to B with respect to First observer which is on the earth(reference S)? Second observer which is in the spaceship (reference S')?
Answer: a) The time needed to travel from A to B with respect to the first observer on S is Lp t, (1) v Where L p is the proper length, i.e, the distance between A and B as measured by the first observer on the earth (reference S). b) The time needed to travel from A to B with respect to the second observer in the spaceship (reference S') is L t', (2) v Where L is the length (distance) as measured by the second observer in the spaceship (reference S').
Dividing (1) and (2) we get t' t L v L v L p L p. (3) Using the relation of time dilation t = γ t, (3) becomes t γ t = 1 γ = L L p Then L p = γl, γ = 1 It is clear that L p < L. 1 v2 c 2 < 1. Definition of length contraction: The length L p (prper length), measured by the first observer on the earth (reference S) is less than the length L measured by the second observer in the spaceship (reference S').
Example: suppose a stick moves past a stationary Earth observer with a speed v=0.95c, as in Figure 1.13b. The length of the stick as measured by an observer in the frame attached to it is the proper length L p =100m as illustrated in Figure 1.13a. What is the length of the stick, L, as measured by the Earth observer? Solution: L is shorter than L p by the factor (1 - v 2 /c 2 ) 1/2. Therefore L=(1 - v 2 /c 2 ) 1/2 L p
= 1 0.95c 2 c 2 100 = 1 0.9025 100 = 0.0975 100 = 0.3123 100 = 31.23m
EXAMPLE A spaceship is measured to be 100 m long while it is at rest with respect to an observer. If this spaceship now flies by the observer with a speed of 0.99c, what length will the observer find for the spaceship? Solution The proper length of the ship is 100 m. From Equation 1.10, the length measured as the spaceship flies by is
EXAMPLE An observer on Earth sees a spaceship at an altitude of 435 m moving downward toward the Earth at 0.970c. What is the altitude of the spaceship as measured by an observer in the spaceship? Solution The proper length here is the Earth ship separation as seen by the Earth-based observer, or 435 m. The moving observer in the ship finds this separation (the altitude) to be
1.6 THE LORENTZ TRANSFORMATION We have seen that the Galilean transformation is not valid when v approaches the speed of light, (v~c) The Lorentz coordinate transformation is a set of formulas that relates the space and time coordinates of two inertial observers moving with a relative speed v. We have already seen two consequences of the Lorentz transformation in the time dilation and length contraction formulas. The Lorentz velocity transformation is the set of formulas that relate the velocity components u x, u y, u z of an object moving in frame S to the velocity components u' x, u' y, u' z of the same object measured in frame S', which is moving with a speed v relative to S.
consider the standard frames, S and S, with S' moving at a speed v along the +x direction (see Fig. 1.2). The origins of the two frames coincide at t' = t = 0. A reasonable guess about the dependence of x on x and t is where G is a dimensionless factor that does not depend on x or t but is some function of v/c such that G is 1 in the limit as v/c approaches 0. Assuming that Equation 1.16 is correct, we can write the inverse Lorentz coordinate transformation for x in terms of x' and t' as
To obtain the inverse Lorentz transformation of any quantity, simply interchange primed and unprimed variables and reverse the sign of the frame velocity v. Derivation of Lorentz Transformations: It is clear that x =ct as measured by S, and x=ct as measured by S, and according to relativity t t. Equations (1.16) and (1.17) becomes ct =G(ct - vt) and ct=g(ct + vt ) Substituting from the second equation into the first and solving for t we get
Taking differentials of Equations 1.16 and 1.18 yields Forming u' x = dx'/dt leads, after some simplification, to where u x = dx/dt. Postulate 2 requires that the velocity of light be c for any observer, so in the case u x = c, we must also have u' x =c. Using this condition in Equation 1.21 gives
Equation 1.22 may be solved to give The direct coordinate transformation is thus x' = (x -vt), and the inverse transformation is x = (x' vt'). To get the time transformation (t' as a function of t and x), substitute G = into Equation 1.18 to obtain
Summary: Lorentz Transformation of coordinate and time from (x, y, z, t) in S to (x', y', z', t') in S' are
where The inverse transformations from (x', y', z', t') in S' to (x, y, z, t) in S are where
Limiting case: When v<<c, Lorentz Transformation Galilean Transformation When v <<c, then v 2 /c 2 << 1, so Lorentz transformation (Equations 1.23 1.26) reduce in this limit to the Galilean coordinate transformation equations, given by
Lorentz Velocity Transformation Substitution of into Equation 1.21, we get where u' x = dx'/dt is the instantaneous velocity in the x direction measured in S and u x = dx /dt is the velocity component of the object as measured in S. Similarly, if the object has velocity components along y and z, the components in S' are
Limiting case: When v<<c, Lorentz velocity Transformation Galilean velocity Transformation When u x v <<c, then v 2 /c 2 << 1, so Lorentz velocity transformation (Equations 1.28 1.29) reduce in this limit to the Galilean coordinatevelocity transformation equations, given by u' x =u x v, u y.=u y, and u y.=u y This corresponds to the Galilean velocity transformation. In the other extreme, when u x = c, Equation 1.28 becomes
Inverse velocity Transformation: To obtain u x in terms of u' x, replace v by v in Equation 1.28 and interchange u x and u' x, this gives EXAMPLE 1.8 Two spaceships A and B are moving in opposite directions, as in Figure 1.19. An observer on Earth measures the speed of A to be 0.750c and the speed of B to be 0.850c. Find the velocity of B with respect to A.
Solution This problem can be solved by taking the S' frame to be attached to spacecraft A, so that v =0.750c relative to an observer on Earth (the S frame). Spacecraft B can be considered as an object moving to the left with a velocity u x = -0.850c relative to the Earth observer. Hence, the velocity of B with respect to A can be obtained using Equation 1.28: The negative sign for u' x indicates that spaceship B is moving in the negative x direction as observed by A. Note that the result is less than c. That is, a body with speed less than c in one frame of reference must have a speed less than c in any other frame. If the incorrect Galilean velocity transformation were used in this example, we would find that u' x = u x - v = -0.850c - 0.750c = -1.600c, which is greater than the universal limiting speed c.
EXAMPLE 1.9 Imagine a motorcycle rider moving with a speed of 0.800c past a stationary observer, as shown in Figure 1.20. If the rider tosses a ball in the forward direction with a speed of 0.700c with respect to himself, what is the speed of the ball as seen by the stationary observer?
Solution : In this situation, the velocity of the motorcycle with respect to the stationary observer is v = 0.800c. The velocity of the ball in the frame of reference of the motorcyclist is u x =0.700c. Therefore, the velocity, u x, of the ball relative to the stationary observer is